15.S24 Sample Exam Solutions

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1 5.S4 Sample Exam Solutions. In each case, determine whether V is a vector space. If it is not a vector space, explain why not. If it is, find basis vectors for V. (a) V is the subset of R 3 defined by 4x 5y + z =. (b) Let the vector w = (w, w,, w n ) represent a portfolio s holdings, where each component w i represents the fraction of the portfolio s total market value in asset i. Let V be the set of weight vectors that can represent market-neutral long/short portfolios. The weights w i satisfy < w i for long positions, w i < for short positions, and w i =. (c) V is the set of vectors in R for which Mv = v, where ( ). 3 i (a) V is a two-dimensional vector space, and a basis is given by any linearly independent vectors satisfying 4x 5y + z =, such as 5 v =, v = 4. (b) V is not a vector space because it is not closed under scalar multiplication or addition, which violate the inequality as well as the budget constraint. (c) M has eigenvalues of and, so the eigenvector ( ) v = is a basis for the vector space V. Summer 4 5.S4 Sample Exam Page

2 . Calculate the trace and the determinant of the matrix. If the matrix is non-singular, compute its inverse. If the matrix is singular, determine its image and kernel. (a) ( ) 3 4 (b) ( ) (c) (d) (e) ( ) 3 ( ) ρ ρ ( ) x x x x (a) (b) (c) ( ), Tr 5, Det, M = ( ) ( ) {( )} {( )} 3 4 4, Tr, Det, Im span, Ker span ( ), Tr, Det, M = 3 ( 3 ) (d) M is non-singular provided ρ. ( ) ρ ρ, Tr, Det ρ, M = ρ ( ) ρ ρ (e) ( x x x x ), Tr, Det, Im span {( )} x, Ker span {( )} x Summer 4 5.S4 Sample Exam Page

3 3. Consider the quadratic form defined by Q(x, y) = 4x + 4xy + y. Find the maxima of Q(x, y) and their location (x, y) subject to constraints as below: (a) Max Q, subject to x + y =. (b) Max Q, subject to x + y = 3. (c) Max Q, subject to x + y =. (a) The Lagrange function is L = Q(x, y) + λ(x + y ), so taking partial derivatives gives three linear equations in three unknowns, = 8x + 4y + λ =, x = 4x + y + λ =, y λ = x + y =. The solution is x = /9, y = 8/9, λ = /9, Q max = /9 (b) The only change is the constant of the constraint equation, λ = x + y 3 =. The solution is x = /3, y = 8/3, λ = /3, Q max = Note that if we define x = 3x and y = 3y, then the constraint is x + y = as above, while the quadratic form scales as Q(x, y) = 9Q(x, y ), so that Q max is 9 times larger than above. (c) With a quadratic constraint, the equations change slightly but become more complex = 8x + 4y + λx =, x = 4x + y + λy =, y λ = x + y =. Summer 4 5.S4 Sample Exam Page 3

4 Multiply the first equation by y and the second by x and then subtract to eliminate λ, leaving So the solution is x + 7xy y = = (3x + 4y)(4x 3y), x + y =. (x, y, λ) = ±(3/5, 4/5, ), Q max =. Summer 4 5.S4 Sample Exam Page 4

5 4. Suppose that a set of portfolio managers has a chance p = 5% of beating the market by % in a given year and chance p of underperforming by %. Performance from one year to the next is independent and uncorrelated. Returns are simple returns and compounding is ignored. (a) What is the probability that a manager will achieve a five-year track record which beats the market for at least 4 out of 5 years? (b) Suppose that unsuccessful managers get forced out of business as soon they are down overall 3%; that is, as soon as as their record contains 3 more losing years than winning years. What is the probability of failure over a five-year horizon? (c) Among those who survive, what is the expected total return? (a) The chance of beating the market at least 4/5 years is 8.75% = 6/3. The chance of 5 winners in a row is /3; the chance of 4 out of 5 is equal to 5/3 since there are 5 different years in which the a single loss could occur; and since these are mutually exclusive, their probabilities add. (b) The probability of ruin is 7/3 = %. This is an example of the absorbing barrier. The track records, or sample paths, that lead to going out of business can be grouped by the number of initial losses that precede the first success. They are F F F xx (4/3) F F SF F (/3) F SF F F (/3) SF F F F (/3) (c) Among the survivors, the expected return will be positive, an effect known as survivorship bias, which is just a consequence of conditional probability. The overall chance and returns are 5 winning years (/3, +5%) 4 winning years (5/3, +3%) 3 winning years (/3, +%) winning years (9/3, -%, since one of the ( 5 ) patterns, F F F SS, is absent because it is not a survivor having lost the first 3 years in a row) Applying the rules of conditional probability, the desired expectation is µ = (/3)(.5) + (5/3)(.3) + (/3)(.) + (9/3)(.) (/3) + (5/3) + (/3) + (9/3) = 8.4% Summer 4 5.S4 Sample Exam Page 5

6 5. Suppose there are 4 assets whose returns have pairwise correlations that are all equal, {, i = j, Corr(r i, r j ) = ρ, i j. Then the correlation matrix is given by ρ ρ ρ C = ρ ρ ρ ρ ρ ρ ρ ρ ρ with eigenvectors v =, v =, v 3 = (a) What are the eigenvalues of C? (b) What is the trace of C? (c) What is the determinant of C?, v 4 = (d) What are the allowed values of ρ for C to be a valid correlation matrix? (Hint: recall that a correlation matrix is always positive semi-definite.) (a) The eigenvalues are λ = + 3ρ and λ j = ρ for j =, 3, 4. (b) Tr C = 4. (c) Det C = λ i = ( + 3ρ)( ρ) 3. (d) Since C is a correlation matrix, the eigenvalues must all be non-negative. So in addition to the usual restriction ρ, we have λ. Therefore 3 ρ.. Summer 4 5.S4 Sample Exam Page 6

7 6. Let X be a random variable with a uniform distribution on the finite interval [, θ], where θ > is unknown. Suppose that a random sample of size n is drawn from the distribution, with observations x,..., x n. Write down the likelihood function for the parameter θ, and find the maximum likelihood estimate (MLE) for θ. The likelihood function is L = (θ + ) n. The likelihood is maximized when (θ + ) is minimized. The possible estimates are constrained by the data, which respects min i {x i } max{x j } θ. So the MLE is θ = max{x j }. j Summer 4 5.S4 Sample Exam Page 7